# How to use the discriminant to find out what type of solutions the equation has for 3z^2 + z - 1 = 0?

$3 {z}^{2} + z - 1$ is of the form $a {z}^{2} + b z + c$, with $a = 3$, $b = 1$ and $c = - 1$.
$\Delta = {b}^{2} - 4 a c = {1}^{2} - \left(4 \times 3 \times - 1\right) = 1 + 12 = 13$
Since $\Delta > 0$, the quadratic has two distinct real roots. However, $\Delta$ is not a perfect square, so those roots are irrational (involving $\sqrt{13}$).